Considering Mach was derived from BSD, and then parts of it like the virtual memory management were integrated back into BSD, it doesn't make much sense to argue that something's not BSD because it's Mach.

"I mean, when you call "android" a linux, then OSX is a BSD. Whether Free or not.

By the logic that Android is Linux, OS X would be Mach.

I'd love to see an analysis of OS X to see just how much BSD derived code there is in there compared to code from other sources. "

Mach doesn't really have a userland. In order to get a working Mach system up & running, the normal approach is to use a BSD userland. So, no matter how you look at it, MacOS X's architecture is consistent with various parts of the BSD tree.